= Solution
Direct calculation of the <curl> gives
$$
\boldsymbol\nabla\times\mathbf v_1=2\mathbf a,\qquad
\boldsymbol\nabla\times\mathbf v_2=\mathbf0.
$$
Since $\mathbf a\ne0$, the first <vector field> cannot be a <gradient field>: the <curl of a gradient> is zero. The second is a smooth <polynomial> field on all of the <simply connected> space $\mathbb R^3$, so its vanishing <curl> makes it a <conservative vector field>. A potential is also constructed explicitly in the next part. Thus \b[$\mathbf v_2$ is a <conservative vector field> and $\mathbf v_1$ is not]. A vanishing <divergence> alone, as for $\mathbf v_1$, is a different property and does not establish conservativity.
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